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From: greg@harvard.UUCP (Greg)
Newsgroups: net.math
Subject: Geodesics on the ellipsoid
Message-ID: <554@harvard.UUCP>
Date: Fri, 13-Dec-85 17:45:54 EST
Article-I.D.: harvard.554
Posted: Fri Dec 13 17:45:54 1985
Date-Received: Sun, 15-Dec-85 07:03:18 EST
Organization: Harvard
Lines: 35
Keywords: elliptic integrals

Somone posted a long time ago asking about geodesics on a circular ellipsoid,
and whether or not they had anything to do with elliptic integral.  The answer
is yes.  Here is how you do it:

I will use the coordinates phi and theta, where:

x = a*cos(phi)*sin(theta)
y = a*sin(phi)*sin(theta)
z = b*cos(theta)

where a does not have to equal b.  Let c^2 = b^2 - a^2.  The metric is given
by ds^2 = dx^2 + dy^2 + dz^2 = (a^2+c^2*cos(theta)^2)*dtheta^2 + a^2*
sin(theta)^2*dphi^2, where s is the distance along the geodesic.  We know
that because rotation about the z-axis is a symmetry, L = a^2*sin(theta)^2
* dphi/ds is a conserved quantity along the geodesic.  We have:

1 = (a^2+c^2*cos(theta)^2)*(dtheta/ds)^2 + a^2*sin(theta)^2*(dphi/ds)^2
= (a^2+c^2*cos(theta)^2)*(dtheta/ds)^2 + (L/a)^2/sin(theta)^2

If u = cos(theta), du = -dtheta * sin(theta), so the above simplifies to:

1 = (a^2 + c^2*u^2)*(du/ds)^2/sin(theta)^2 + (L/a)^2/sin(theta)^2, or

1-u^2 = (a^2 + c^2*u^2)*(du/ds)^2 + (L/a)^2.

With a little more algebra, one gets:

du*(a^2+c^2*u^2)/sqrt((1-(L/a)^2-u^2)*(a^2+c^2*u^2)) = ds

For the solution, then, we must integrate both sides.  I think that the right
side can be reduced to an elliptic integral, but since elliptic integrals are
really messy, and since I don't know them very well, I don't care to work out
the solution any farther.
-- 
gregregreg